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Python

Asher Orr
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Asher Orr
Python Development Techdegree Graduate 9,408 Points
Posted by Asher Orr
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Asher Orr
Python Development Techdegree Graduate 9,408 Points

Not receiving "{} is neither a fizzy or buzzy number" when entering numbers not divisible by 3 or 5.

Hi everyone! I created a function to calculate whether a number is a FizzBuzz, a Fizz, a Buzz, or neither fizzy or buzzy.

Here's my code:

def fizz_buzz_calculator(number):
    if number % 3 == 0 and number % 5 == 0:
         print("{} is a FizzBuzz number.".format(number))
    elif number % 3==0:
        print("{} is a Fizz number.".format(number))
    elif number % 5==0:
         print("{} is a Buzz number.".format(number))
    elif number % 3 == 1 and number % 5 == 1:
         print("{} is neither a fizzy or a buzzy number.".format(number))

name = input("Please enter your name: ")
try:
    number = int(input("Please enter a number: "))
except ValueError:
    print("Please enter only a whole number.")
else:
    print("Hi, {}!".format(name))
    print("You entered the number {}.".format(number))
    fizz_buzz_calculator(number)

If I understand correctly, my function is saying:

  • if the number is evenly divisible by 3 and 5, print "is a Fizzbuzz number."
  • if that's not true, try dividing the number by 3. If it's evenly divisible by 3, print "is a fizz number."
  • if that's not true, try dividing the number by 5. If it's evenly divisible by 5, print "is a buzz number."
  • If none of the above arguments are true, try seeing if the number is not evenly divisible by 3 and 5. If that's the case, print "{} is neither a fizzy or a buzzy number."

When I pass numbers like 15, 5, and 6, my code works. But when I pass a number that isn't evenly divisible by either 3 or 5, like the number 8, my function doesn't run.

The last thing I see in the console is "You entered the number {}."

Can anyone help me here? Thanks so much for reading!

EDIT:

Hello again! Shortly after posting the question above, I updated my code:

def fizz_buzz_calculator(number):
    if number % 3 == 0 and number % 5 == 0:
         print("{} is a FizzBuzz number.".format(number))
    elif number % 3==0:
        print("{} is a Fizz number.".format(number))
    elif number % 5==0:
         print("{} is a Buzz number.".format(number))
    else:
         print("{} is neither a fizzy or a buzzy number.".format(number))

name = input("Please enter your name: ")
try:
    number = int(input("Please enter a number: "))
except ValueError:
    print("Please enter only a whole number.")
else:
    print("Hi, {}!".format(name))
    print("You entered the number {}.".format(number))
    fizz_buzz_calculator(number)

It's working now, but I'm confused as to why the original function did not work. I would appreciate it if someone could explain to me what was happening! Thank you!

2 Answers

jb30
jb30
44,806 Points

number % 3 == 1 and number % 5 == 1 will be True when number is 1 more than a multiple of 15, and False otherwise.

There are three possibilities for number % 3: 0, 1, and 2. There are five possibilities for number % 5: 0, 1, 2, 3, and 4. The previous elif statements dealt with the cases of number % 3 == 0 and number % 5 == 0, so in the final elif statement, number % 3 will be 1 or 2 and number % 5 will be 1, 2, 3, or 4.

When number is 8, 8 % 3 is 2, which is not equal to 1, so the condition is False and nothing gets printed.

When number is 7, 7 % 3 is 1, which is equal to 1, but 7 % 5 is equal to 2, which is not equal to 1. True and False becomes False, so nothing gets printed.

When number is 16, 16 % 3 is 1 and 16 % 5 is 1. 1 == 1 and 1 == 1 is True, so the original code will print 16 is neither a fizzy or a buzzy number.

Steven Parker
Steven Parker
231,007 Points
Steven Parker
Steven Parker
231,007 Points

Your final test in the first example is elif number % 3 == 1 and number % 5 == 1 and neither of those conditions would be true for the number 8. It would work only for numbers like 1, 16, 31, 46 … etc., and that's why you didn't see the final message.

The important thing is that you realized that you needed just a plain "else" to handle all numbers not previously identified.